(1)
证明:连接AD.
∵AB是直径,∴∠ACB=∠ADB=90°.
∵AC=BD,AB=BA,∴△ABC≌△ABD.
∴∠BAC=∠ABD,从而PA=PB.
∵O是AB中点,∴PO⊥AB;
(2)∵∠AOP=∠ACB=90°,∠OAP=∠CAB,
∴△AOP∽△ACB
∴
.
∵AB=4,BC=1,
∴AC=
=
.
∴OP=
=
.
(1)
证明:连接AD.
∵AB是直径,∴∠ACB=∠ADB=90°.
∵AC=BD,AB=BA,∴△ABC≌△ABD.
∴∠BAC=∠ABD,从而PA=PB.
∵O是AB中点,∴PO⊥AB;
(2)∵∠AOP=∠ACB=90°,∠OAP=∠CAB,
∴△AOP∽△ACB
∴
.
∵AB=4,BC=1,
∴AC=
=
.
∴OP=
=
.